Why $\sqrt{\rho} = P \sqrt{\rho}$ in the proof of quantum error correction conditions in Nielsen & Chuang?

I have trouble understanding a proof in Nielsen & Chuang, specifically the identity in (10.20), which reads $$U_k^\dagger P_k F_l \sqrt{\rho} = U_k^\dagger P_k^\dagger F_l P \sqrt{\rho}.$$

By playing around with the definitions, the only way how I would get the rightmost $$P$$ to appear would be by assuming $$\sqrt{\rho} = P \sqrt{\rho}$$. In this answer it is stated that "$$P$$ is the projector onto the code space and thus does nothing to $$\rho$$". But I don't get this statement.

Recall from the sentence immediately preceding $$(10.20)$$ that $$\rho$$ is a state in the code subspace $$C$$ and from the statement of theorem $$10.1$$ that $$P$$ is a projector onto $$C$$. We will show that $$\sqrt{\rho}=P\sqrt{\rho}$$.

Let $$|i_L\rangle$$ be an orthonormal basis of the code subspace. Then $$P=\sum_i|i_L\rangle\langle i_L|.\tag1$$ Therefore, for any $$|j_L\rangle$$, we have$$^1$$ $$P|j_L\rangle=\sum_i|i_L\rangle\langle i_L|j_L\rangle=|j_L\rangle.\tag2$$ Now, if $$\rho$$ has eigendecomposition $$\rho=\sum_kp_k|\psi_k\rangle\langle\psi_k|\tag3$$ then $$\sqrt{\rho}=\sum_k\sqrt{p_k}|\psi_k\rangle\langle\psi_k|.\tag4$$ But $$\rho$$ is in the code subspace, so $$|\psi_k\rangle=\sum_i\alpha_{ki}|i_L\rangle$$ for some $$\alpha_{ki}\in\mathbb{C}$$. Substituting into $$(4)$$, we get $$\sqrt{\rho}=\sum_{ijk}\sqrt{p_k}\alpha_{ki}\overline{\alpha_{kj}}|i_L\rangle\langle j_L|.\tag5$$ Finally, calculate \begin{align} P\sqrt{\rho}&=P\sum_{ijk}\sqrt{p_k}\alpha_{ki}\overline{\alpha_{kj}}|i_L\rangle\langle j_L|\tag6\\ &=\sum_{ijk}\sqrt{p_k}\alpha_{ki}\overline{\alpha_{kj}}P|i_L\rangle\langle j_L|\tag7\\ &=\sum_{ijk}\sqrt{p_k}\alpha_{ki}\overline{\alpha_{kj}}|i_L\rangle\langle j_L|\tag8\\ &=\sqrt{\rho}\tag9 \end{align} where we used $$(5)$$, linearity of $$P$$, $$(2)$$, and finally $$(5)$$ again.

$$^1$$ Equivalently, one may take $$(2)$$ together with the requirement that $$P$$ vanishes on vectors orthogonal to the space spanned by $$|i_L\rangle$$ as the definition of the projector.

• Very helpful, thanks! Commented Dec 1, 2023 at 16:21

$$\rho$$ is a state in the code. $$P$$ is the projector onto the codespace. That means (by definition) that for any state $$|\psi\rangle$$ in the code, $$P|\psi\rangle=|\psi\rangle$$.

$$\rho$$ is one such state (except that it's possibly mixed), so it must satisfy $$P\rho=\rho$$. Thus, if $$\rho$$ appears somewhere in an equation, you can arbitrarily choose to replace it with $$P\rho$$ if you want. Furthermore, if $$\rho$$ is in the code, all its eigenvectors are in the code. Hence $$\sqrt{\rho}$$, which has the same eigenvectors, must also be in the code.

• Thanks, but shouldn't it be $P \rho P = \rho$? Commented Dec 1, 2023 at 16:25
• That is also true. Think about a pure state $P|\psi\rangle=|\psi\rangle$. So, if I had $\rho=|\psi\rangle\langle\psi|$, then it is true that $\rho=P\rho=\rho P=P\rho P$ Commented Dec 1, 2023 at 16:31